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Schedule…

Schedule…. Divine Source. 2 Nephi 25:26   26 And we talk of Christ, we rejoice in Christ, we preach of Christ, we prophesy of Christ, and we write according to our prophecies, that our children may know to what source they may look for a remission of their sins. .

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Schedule…

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  1. Schedule… Discussion #2 – Kirchhoff’s Laws

  2. Divine Source 2 Nephi 25:26   26 And we talk of Christ, we rejoice in Christ, we preach of Christ, we prophesy of Christ, and we write according to our prophecies, that our children may know to what source they may look for a remission of their sins. Discussion #2 – Kirchhoff’s Laws

  3. Lecture 2 – Kirchhoff’s Current and Voltage Laws Discussion #2 – Kirchhoff’s Laws

  4. Charge • Elektron: Greek word for amber • ~600 B.C. it was discovered that static charge on a piece of amber could attract light objects (feathers) • Charge (q): fundamental electric quantity • Smallest amount of charge is that carried by an electron/proton (elementary charges): Coulomb (C): basic unit of charge. Discussion #2 – Kirchhoff’s Laws

  5. i Area Electric Current • Electric current (i): the rate of change (in time) of charge passing through a predetermined area (IE the cross-sectional area of a wire). • Analogous to volume flow rate in hydraulics • Current (i) refers to ∆q(dq) units of charge that flow through a cross-sectional area (Area) in ∆t(dt)units of time Ampere (A): electric current unit. 1 ampere = 1 coulomb/second (C/s) Positive current flow is in the direction of positive charges (the opposite direction of the actual electron movement) Discussion #2 – Kirchhoff’s Laws

  6. Charge and Current Example • For a metal wire, find: • The total charge (q) • The current flowing in the wire (i) • Given Data: • wire length = 1m • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Discussion #2 – Kirchhoff’s Laws

  7. Charge and Current Example • For a metal wire, find: • The total charge (q) • The current flowing in the wire (i) • Given Data: • wire length = 1m • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Discussion #2 – Kirchhoff’s Laws

  8. Charge and Current Example • For a metal wire, find: • The total charge (q) • The current flowing in the wire (i) • Given Data: • wire length = 1m • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Discussion #2 – Kirchhoff’s Laws

  9. Charge and Current Example • For a metal wire, find: • The total charge (q) • The current flowing in the wire (i) • Given Data: • wire length = 1m • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Discussion #2 – Kirchhoff’s Laws

  10. i i1 i3 i2 + 1.5 V _ i Kirchhoff’s Current Law (KCL) • KCL: charge must be conserved – the sum of the currents at a node must equal zero. Node 1 At Node 1: -i + i1 + i2 + i3 = 0 OR: i - i1 - i2 - i3 = 0 NB: a circuit must be CLOSED in order for current to flow Discussion #2 – Kirchhoff’s Laws

  11. Node 1 i i1 i6 i2 i3 i4 i5 + 1.5 V _ i Kirchhoff’s Current Law (KCL) • Potential problem of too many branches on a single node: • not enough current getting to a branch • Suppose: • all lights have the same resistance • i4 needs 1A • What must the value of i be? Discussion #2 – Kirchhoff’s Laws

  12. Node 1 i i1 i6 i2 i3 i4 i5 + 1.5 V _ i Kirchhoff’s Current Law (KCL) • Potential problem of too many branches on a single node: • not enough current getting to a branch -i + i1 + i2 + i3 + i4 + i5 + i6 = 0 BUT: since all resistances are the same: i1 = i2 = i3 = i4 = i5 = i6 = in -i + 6in= 0 6in = i 6(1A) = i i = 6A Discussion #2 – Kirchhoff’s Laws

  13. + Vs _ Kirchhoff’s Current Law (KCL) • Example1: find i0 and i4 • is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A i0 i1 i2 is i3 i4 Discussion #2 – Kirchhoff’s Laws

  14. + Vs _ Kirchhoff’s Current Law (KCL) • Example1: find i0 and i4 • is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A Node a • NB: First thing to do – decide on unknown current directions. • If you select the wrong direction it won’t matter • a negative current indicates current is flowing in the opposite direction. • Must be consistent • Once a current direction is chosen must keep it i0 i1 i2 Node b is i3 i4 Node c Discussion #2 – Kirchhoff’s Laws

  15. Node a i0 i1 i2 Node b + is Vs i3 _ i4 Node c Kirchhoff’s Current Law (KCL) • Example1: find i0 and i4 • is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A Discussion #2 – Kirchhoff’s Laws

  16. + _ Vs2 + _ Vs1 Kirchhoff’s Current Law (KCL) • Example2: using KCL find is1 and is2 • i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A is2 is1 i2 R2 R4 i4 R3 i5 i3 R5 Discussion #2 – Kirchhoff’s Laws

  17. is2 Supernode is1 i2 + R2 _ Vs2 + _ Vs1 R4 i4 R3 i5 i3 R5 Kirchhoff’s Current Law (KCL) • Example2: using KCL find is1 and is2 • i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A Discussion #2 – Kirchhoff’s Laws

  18. Node a is2 is1 i2 + R2 _ Vs2 + _ Vs1 R4 i4 R3 i5 i3 R5 Kirchhoff’s Current Law (KCL) • Example2: using KCL find is1 and is2 • i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A Discussion #2 – Kirchhoff’s Laws

  19. Voltage • Moving charges in order to produce a current requires work • Voltage: the work (energy) required to move a unit charge between two points • Volt (V): the basic unit of voltage (named after Alessandro Volta) Volt (V): voltage unit. 1 Volt = 1 joule/coulomb (J/C) Discussion #2 – Kirchhoff’s Laws

  20. vba _ + a b _ + vab Voltage • Voltage is also called potential difference • Very similar to gravitational potential energy • Voltages are relative • voltage at one node is measured relative to the voltage at another node • Convenient to set the reference voltage to be zero vab=> the work required to move a positive charge from terminal a to terminal b vba=> the work required to move a positive charge from terminal b to terminal a vba = - vab vab = va - vb Discussion #2 – Kirchhoff’s Laws

  21. i i a b a b _ _ + + vba vab Voltage • Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied • Since i is going from + to – energy is being absorbed by the element (passive element) • Since i is going from – to + energy is being supplied by the element (active element) Discussion #2 – Kirchhoff’s Laws

  22. i i a i b a b _ _ a + + vba vab + + + vab v1 1.5 V _ _ _ i b Voltage • Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied Absorbing energy (load) (passive element) POSITIVE voltage Supplying energy (source) (active element) NEGATIVE voltage Discussion #2 – Kirchhoff’s Laws

  23. Voltage • Ground: represents a specific reference voltage • Most often ground is physically connected to the earth (the ground) • Convenient to assign a voltage of 0V to ground The ground symbol we’ll use (earth ground) Another ground symbol (chasis ground) Discussion #2 – Kirchhoff’s Laws

  24. i a + + + vab v1 1.5 V _ _ _ i b Kirchhoff’s Voltage Law (KVL) • KVL: energy must be conserved – the sum of the voltages in a closed circuit must equal zero. Use Node b as the reference voltage (ground): vb = 0 Discussion #2 – Kirchhoff’s Laws

  25. + v2 – + _ Vs1 i + v1 – + v3 – Kirchhoff’s Voltage Law (KVL) • Example3: using KVL, find v2 • vs1 = 12V, v1 = 6V, v3 = 1V • Source: loop travels from – to + terminals • Sources have negative voltage • Load: loop travels from + to – terminals • Loads have positive voltage Discussion #2 – Kirchhoff’s Laws

  26. + v2 – + _ Vs1 i + v1 – + v3 – Kirchhoff’s Voltage Law (KVL) • Example3: using KVL, find v2 • vs1 = 12V, v1 = 6V, v3 = 1V • Source: loop travels from – to + terminals • Sources have negative voltage • Load: loop travels from + to – terminals • Loads have positive voltage Discussion #2 – Kirchhoff’s Laws

  27. + v2 – + _ Vs1 i + v1 – + v3 – Kirchhoff’s Voltage Law (KVL) • Example3: using KVL, find v2 • vs1 = 12V, v1 = 6V, v3 = 1V NB: v2 is the voltage across two elements in parallel branches. The voltage across both elements is the same: v2 Discussion #2 – Kirchhoff’s Laws

  28. + Vs2 _ + Vs1 _ Kirchhoff’s Voltage Law (KVL) • Example4: using KVL find v1 and v4 • vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V + v1 – + v2 – + v4 – + v3 – + v5 – Discussion #2 – Kirchhoff’s Laws

  29. + Vs2 _ + Vs1 _ Kirchhoff’s Voltage Law (KVL) • Example4: using KVL find v1 and v4 • vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V + v1 – + v2 – Loop1 Loop2 + v4 – + v3 – + v5 – Loop3 Discussion #2 – Kirchhoff’s Laws

  30. + v1 – + v2 – + Vs2 _ + Vs1 _ + v4 – + v3 – + v5 – Kirchhoff’s Voltage Law (KVL) • Example4: using KVL find v1 and v4 • vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V Loop1 Loop2 Loop3 Discussion #2 – Kirchhoff’s Laws

  31. + v1 – + v2 – + Vs2 _ + Vs1 _ + v4 – + v3 – + v5 – Kirchhoff’s Voltage Law (KVL) • Example4: using KVL find v1 and v4 • vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V Loop1 Loop2 Loop3 Discussion #2 – Kirchhoff’s Laws

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